MATH 412 PROBLEM 8 SOLUTIONS

Reading: • Hungerford All of Chapter 7 and 8.1

Practice Problems: These are “easier” problems critical for your understanding, but not to be turned in. Be sure you can do these! You may see them or something very similar on Quizzes and Exams. • 7.1: 25, • 7.2: 14 • 7.3: 11, 12, 19 • 7.4: 29, 31, 40-44. • 7.5: 11, 31 • 8.1: 15

Problems due Friday. Write out carefully, staple pages for A and B together. Staple pages for C and D together separately. Write your name and section number on each.

A. Let S1 be the subset of C consisting of of absolute value 1; that is 1 S := {z ∈ C | |z| = 1}. (1) Prove that S1 is a of C×. (2) Prove that the map x −y S1 → SL ( ) x + iy 7→ 2 R y x is an injective . 1 (3) Prove that S is isomorphic to SO2(R), the group of orthogonal matrices of determinant 1. Use this to give a geometric interpretation of the group S1 that explains why some call it the “continuous rotation group." (4) For every positive n, find an element of n in S1. (5) Find an element of infinite order in S1. (6) For each positive integer n, find a subgroup of S1 of order n.

Solution. (1) Clearly, since 1 ∈ S1, we see S1 is non-empty. Also, if z, w ∈ S1, then zw ∈ S1, since |zw| = |z||w| = 1, so S1 is closed under multiplication. Finally, if z ∈ S1, then since zz−1 = 1, we know |z||z−1| = |1| = 1. This tells us that |z−1| = 1, so S−1 is closed under taking inverses. Thus S1 is a subgroup of C×. x −y a −b (2) Since (x + iy)(a + bi) = (xa − yb) + i(ya + xb) We need to check that = y x b a (xa − yb) −(ya + xb) . This is true. (ya + xb)(xa − yb)

1 MATH 412 PROBLEM SET 8 SOLUTIONS 2

1 (3) The map in (2) defines an from S to the in GL2(R). To see the image is in SO2(R), note simply that the columns are orthonormal and the determinant is 1 since x2 + y2 = 1 by virtue of being in S1. cos(2π/n) − sin(2π/n) (4) The matrix is order n, hence so is cos(2π/n) + i sin(2π/n) ∈ S1. sin(2π/n) cos(2π/n) (5) For an element of infinite order, take an irrational multiple of 2π. (6) The subgroup generated by the order n element cos(2π/n) + i sin(2π/n) has order n.

B. Consider the group GL2(Zp) of invertible 2 × 2 matrices with entries in the field Zp. This is the group of units in M2(Zp). a a b (1) Prove that for any non-zero column , the matrix ∈ M ( ) is a non- if and only if c c d 2 Zp a b there exists n ∈ such that n = . Zp c d 1 (2) Compute the order of GL2(Zp). 2 (3) Show that the subset of diagonal matrices in GL2(Zp) is an abelian subgroup of order (p − 1) . 2 (4) Show that the upper triangular matrices in GL2(Zp) form a non-abelian subgroup of order p(p−1) . (5) Find a subgroup of order p − 1 in GL2(Zp). 2 (6) Show that the subgroup SL2(Zp) of matrices of determinant 1 has index p − 1 in GL2(Zp). (7) Prove that two matrices A, B in GL2(Zp) are in the same of SL2(Zp) in GL2(Zp) if and only if det A = det B.

Solution. (1) A matrix is a unit if and only if it is invertible, which happens if and only if its columns are linearly independent (from 217). Since we are assuming the first column is non-zero, the matrix fails to be invertible if and only if the second column is a multiple of the first. (2) There are p2 − 1 choices for the first column, and p2 − p for the second, using (1). So 2 2 2 |GL2(Zp)| = (p − 1)(p − p) = p(p + 1)(p − 1) . (3) For the diagonal matrices, we need to have non-zero elements on the diagonal (or the deter- minant would be zero). There are (p − 1)2 of these. We check that they commute with each other by direct computation. (4) The upper triangular matrices again must have non-zero entries on the diagonal (since the 2 determinant is non-zero) but the upper right entry can be arbitrary. So |U2(Zp)| = (p − 1) p. 1 1 1 1 1 1 1 1 To see it is non-abelian, check that 6= . 0 2 0 1 0 1 0 2 a 0 (5) The invertible scalar matrices { | a ∈ ×} are a subgroup of order p − 1. 0 a Zp × (6) There is a surjective homomorphism GL2(Zp) → Zp sending each matrix to its determinant. ∼ Its is the subgroup SL2(Zp). By the first isomorphism theorem, GL2(Zp)/SL2(Zp) = × × Zp . So the number of equals the order of Zp , which is p − 1. So the index is p − 1. (7) Suppose that A, B are in the same (right) coset of SL2(Zp). This means that A = CB where C ∈ SL2(Zp). So det(A) = det(BC) = det(B) det(C) = det(B). Conversely, if det A = det B, −1 −1 then AB has determinant one, so is in SL2(Zp). Let C = AB . Then A = CB, where

1Hint: You might want to use (1) in the course of doing this. 2You can do this one after you do (7) if you like. MATH 412 PROBLEM SET 8 SOLUTIONS 3

C ∈ SL2(Zp). This means that A and B are in the same (right) coset of SL2(Zp). [Note that SL2(Zp) is normal in GL2(Zp) so its left and right cosets are the same.]

C. Consider the following elements in GL2(C): 1 0 i 0   0 1 0 i 1 = , i = , j = , k = , 0 1 0 −i −1 0 i 0

Let Q be the subgroup of GL2(C) generated by the matrices i, j, k. (1) Show3 that Q contains the 8 elements {±1, ±i, ±j, ±k}. (2) Make a multiplication table for Q. (3) Find (with proof) the complete list of all cyclic of Q of order 4. (4) Find (with proof) the complete list of all cyclic subgroups of Q of order 2. (5) Prove that every proper subgroup of Q is cyclic. (6) Can Q be generated by two elements? Prove it. (7) Is Q8 isomorphic to D4? Prove or disprove.

Solution. (1) ◦ 1 −1 i −i j −j k −k 1 1 −1 i −i j −j k −k −1 −1 1 −i i −j j −k k i i −i −1 1 k −k −j j −i −i i 1 −1 −k k j −k j j −j −k k −1 1 i −i −j −j j k −k 1 −1 −i i k k −k j −j −i i −1 1 −k −k k −j j i −i 1 −1 (2) Identifying cyclic subgroups of order 4 amounts to finding elements of order 4. Note that ±i, ±j, ±k are all order 4, and −1 is order 2. So there are there cyclic groups of order 4, namely, those generated by i, j or k. Note that hii = {i, −1, −i, 1}, and similarly for j and k. (3) There is exactly one subgroup of order 2, namely {±1}. (4) The proper subgroups are order 1, 2, and 4 by Lagrange’s theorem. We have listed all the non-trivial ones in the preceding two problems. All are cyclic. (5) Yes, Q = hi, ji. Note that if we have i and j, we have ij = k, and together the powers of i, j, andk yield the rest of the elements of Q. (6) No. D4 has order eight, but it has five element of order 2 (the four reflecttions and rotation through 180). The group Q has one element of order 2.

D. A regular n-gon is a planar figure with n vertices and n congruent sides, with all interior angles congruent. For example, a regular 3-gon is another for an equilateral triangle and a regular 4-gon is the same as a . Let Dn be the of a regular n-gon. To fix ideas, fix a regular n-gon in the Cartesian plane so that its vertices are equidistant from the origin and one lies on the x-axis. It is a fact that |Dn| = 2n. You may use this without proof for this problem (though it may be instructive to think about why this is so).

3It might be helpful to do (2) more or less at the same time. MATH 412 PROBLEM SET 8 SOLUTIONS 4

(1) The group Dn contains an element r of order n. Describe it. (2) For each vertex v of the regular n-gon, Dn contains an of order 2. Describe it. (3) Show that for n odd, the reflections in (2) give n different elements of Dn but if n is even, they give only n/2 different elements of Dn. (4) If n is even, find n/2 reflections in Dn not already described in (3). 4 (5) Show that Dn can be generated by one rotation and one reflection. 5 (6) Show that there is a Dn → {±1} sending each rotation to 1 and each reflection to −1. Be sure to say what group structure you are using on the set {±1}. (7) Find the kernel of the map in (6). Prove that it is a cyclic subgroup. (8) Let R be the subgroup of all rotations of the n-gon. How many cosets are there of R in Dn? Describe the elements of each coset geometrically.

Solution.

(1) Rotation through 2π/n is in Dn. (2) Reflection over the line through v and the origin. (3) If n is odd, the line though v and origin does not pass through another vertex, so there are n different reflections. In n is even, the line we reflect over passes through another vertex, so these vertices determine the same reflection. (4) For n even, we can take the line bisecting an edge of the n-gon (it will necessarily also bisect the opposite side as well). There are n/2 distinct such lines, hence n/2 reflections. (5) The subgroup generated by r has order n. So, given that |Dn| = 2n, we know that, because no reflection is a rotation, the group hr, si, where s is a reflection, is strictly large than hri. But by Lagrange’s theorem, the order of hr, si divides 2n. The only number larger than n that divides 2n is 2n. So hr, si = Dn. (6) Here, consider {±1} as a group under multiplication. Send each of the n reflections to −1 and each of the rotations (including rn = e) to 1. It is easy to see that a composition of two rotations is a rotation, a composition of two reflections is a reflection, and a composition of a rotation and a reflection is a reflection. One one to do this is to write down the matrices, with real coefficients, of each of these transformations (from 217). The rotations all have determinant one and the reflections all have determinant -1. The map is the determinant map. (7) The kernel is hri, the cyclic group generated by rotation through 2π/n. (8) There are two cosets, the rotations are one. The reflections are the other.

4 Use Lagrange’s theorem and the fact you are given that |Dn| = 2n. It is possible to do this without assuming |Dn| = 2n, if you would like a challenge! 5If you get stuck proving this is homomorphism, there are many possible hints I can give. There is a geometric argument in which you can distinguish rotations and reflections by whether they change the orientation. That is, label the vertices 1, 2, . . . , n of the n-gon, so that we go counterclockwise from 1 to n. After a rotation, the vertices still go counterclockwise from 1 to n; after a reflection, they go clockwise from 1 to n. A completely different approach, which is more algebraic, is to use (5). If your rotation in (5) is called r and your reflection is called s, then start writing out the “words" in r and s: are r, r2, . . . , rn−1, e all distinct? what about sr, sr2, . . . , srn−1, s? Write out the “words" in an organized way so that you know −1 n−1 you’ve got each element of Dn on the list exactly once. You might also find it helpful to verify that rs = sr = sr .